If the direction ratios of two lines are give | vedclass

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(D) Given equations are: $3lm - 4ln + mn = 0$ ... $(i)$ $l + 2m + 3n = 0$ ... (ii) From (ii),$l = -(2m + 3n)$. Substituting this in $(i)$: $-3(2m + 3n

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$\frac{\pi}{2}$

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(D) Given equations are: $3lm - 4ln + mn = 0$ ... $(i)$ $l + 2m + 3n = 0$ ... (ii) From (ii),$l = -(2m + 3n)$. Substituting this in $(i)$: $-3(2m + 3n)m + 4(2m + 3n)n + mn = 0$ $-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$ $-6m^2 + 12n^2 = 0$ $m^2 = 2n^2 \Rightarrow m = \pm \sqrt{2}n$. Case $1$: $m = \sqrt{2}n$. Then $l = -(2\sqrt{2} + 3)n$. Direction ratios $(l_1, m_1, n_1) = (-(3 + 2\sqrt{2}), \sqrt{2}, 1)$. Case $2$: $m = -\sqrt{2}n$. Then $l = -(-2\sqrt{2} + 3)n = (2\sqrt{2} - 3)n$. Direction ratios $(l_2, m_2, n_2) = (2\sqrt{2} - 3, -\sqrt{2}, 1)$. For angle $	heta$ between lines: $\cos 	heta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$ Numerator: $(-(3 + 2\sqrt{2}))(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1)$ $= -((2\sqrt{2})^2 - 3^2) - 2 + 1 = -(8 - 9) - 1 = 1 - 1 = 0$. Since the numerator is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is:

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